Let $f(x) = \frac{x^3 + 1}{x}. $ Plot the function using derivatives criteria.
First of all, $f$ is indeterminate at $x=0.$ So I would have to deal with 2 intervals, separately: $(- \infty, 0)$, and $(0, \infty)$.
I would also have to consider:
$\lim_{\to \infty} f(x) = \lim_{\to \infty} \frac{1}{x} + x^2 = 0 + \infty = \infty$
$\lim_{\to -\infty} f(x) = \lim_{\to -\infty} \frac{1}{x} + x^2 = 0 + (-\infty)^2 = \infty$
If $x\neq 0, \frac{x^3 + 1}{x} = 0 \implies x^3 + 1 = 0,$ so the only root in $\mathbb{R}$ is $x=-1.$
If $x>0, \lim_{x \to 0} \frac{x^3 + 1}{x} = ( \lim_{x \to 0} \frac{1}{x} ) ( \lim_{x \to 0} x^3 + 1) = (\infty)(1) = \infty $
And if $x <0, \lim_{x \to 0} \frac{x^3 + 1}{x} = ( \lim_{x \to 0} \frac{1}{x} ) ( \lim_{x \to 0} x^3 + 1) = (-\infty)(1) = -\infty $
Then $f '(x) = \frac{2x^3 -1}{x^2}$
It should happen that, at $(\infty, 0)$, $f$ is decreasing. And $x<0 \implies 2x^3 -1 < 0 \implies \frac{2x^3 -1}{x^2} < 0.$ So $f'(x) < 0 \implies f$ is decreasing at $(-\infty, 0)$
and it should also happen that, at $(0, \infty)$, the function is increasing.
But, if $x<0$, it doesn't neccessarily happen that $2x^3 -1 > 0,$ So I am stuck here. However if i could assume that $2x^3 -1 > 0, $ then $f '(x) = \frac{2x^3 -1}{x^2} > 0,$ so $f$ would be increasing.
To find critical points, $f '(x) = \frac{2x^3 -1}{x^2} = 0 \implies_{x \neq 0} x = \frac{1}{\sqrt[3]{2}},$ or $x = -\frac{1}{\sqrt[3]{2}} ,$ and $f''(x) = \frac{2x^3+2}{x^3},$ so $f''(\frac{1}{\sqrt[3]{2}}) = 4(\frac{1}{2} + 1) = 6 >0,$ that would mean that $f$ has a local minimum. So my last question is why doesn't $x = -\frac{1}{\sqrt[3]{2}} $, the other critical point appears in the graph?
A qualitative approach.
The function can be written as $x^2+\dfrac1x$, and is the sum of a parabola and an equilateral hyperbola.
The hyperbola has a vertical asymptote at $x=0$ and the horizontal asymptote $y=0$. Hence for small $|x|$ it is dominant, and for large $|x|$, it is neglectible and the function gets closer and closer to the parabola.
In the negatives, both functions are decreasing and the sum is decreasing (to $-\infty$).
In the positives we sum increasing and decreasing and we cannot conclude about the variation. The first derivative is $2x-\dfrac1{x^2}$, which has a single positive root, hence the function has a single minimum.
This is confirmed by a plot: